Consider the complete tripartite graph $G = \langle X\cup Y\cup Z,E\rangle$ with $|X|=|Y|=|Z|=\frac{n}{3}$, and $E = X\times Y\cup X\times Z \cup Y\times Z$. This graph is strongly (n,2n/3,n/3,2n/3)-regular, and e.g. $X$ has more than $\frac{n}{4}$ vertices and less than $\frac{3}{4}n$ neighbors, so this is a counterexample to your conjecture.

On the other hand, since the spectrum of strongly regular graphs is known precisely (see https://en.wikipedia.org/wiki/Strongly_regular_graph#Eigenvalues), one can get bounds on the edge expansion of such a graph via Cheeger's inequality, which implies results like the one you conjecture.

Question: What is the minimal $\alpha>0$ such that in every strongly regular graph, every set $S \subseteq V$ whose size is at least $\alpha n$ has at least $(1-\alpha) n$ neighbors, and which strongly regular graph achieves this $\alpha$?

This seems like a natural question, but I wouldn't be surprised if no one had considered it. It looks like a good research question to me. My example implies $\alpha \geq 1/3$.